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Differentiability at a point (old)

An older video where Sal finds the points on the graph of a function where the function isn't differentiable. Created by Sal Khan.
Video transcript
Consider f which is defined for all real numbers. At what arguments x is f of x not differentiable? So to think about that, I'm actually going to try to visualize what f prime of x must look like. So I'm going to do f prime of x in this purple color. So if we look at f of x right over here, it looks like its slope is pretty much consistently negative 2 over this interval between x equals, I guess, it's like negative 8 and 1/2 all the way up to x equals negative 2. It looks like the slope is a constant negative 2. So if I were to draw its derivative, its derivative would look something like this. Its derivative looks something like this. But then something interesting happens at x equals negative 2. Right as we cross x equals negative 2, it looks like the slope goes from being negative to being positive. And it looks like right out the get go, if I were to estimate the slope of its tangent line, it starts changing. It's not a line anymore. It's a curve. The slope of the tangent line right at this point looks like it's around-- I don't know-- it looks like it's around 3 and 1/2. Because if I were to draw a tangent line right over here, it looks like if I move 1 in the x direction, I move up about 3 and 1/2 in the y direction. So I'm just trying to, obviously, estimate it. So it looks like the slope goes up to 3 and 1/2 right when I cross that point. And then the slope becomes lower and lower and lower all the way until I get to this point right over here, all the way until I get to x equals 2. And it looks like it continues to get lower all the way until you get to x equals 3. So it looks like the slope of the line is-- it looks like it's getting lower at a constant rate, I guess I could say. So it looks like it's doing something like this over this interval. But then right as x crosses 3, this becomes a flat line. The slope is 0 here. So right as x crosses 3, the slope becomes 0. So we immediately see there are points where it looks like the slope jumps. And at these points we really don't have a defined derivative. The slope jumps there as well. And so at what arguments is f not differentiable? Well, it's not differentiable when x is equal to negative 2. When x is equal to negative 2, we really don't have a slope there. Remember, when we're trying to find the slope of the tangent line, we take the limit of the slope of the secant line between that point and some other point on the curve. If we did that as we approached from the left, it looks like the derivative is negative 2. If we do that from the right, it looks like the derivative is something like positive 3 and 1/2. And so we're not getting the same limit of the secant line as we approach from the left and as we approach from the right. And the same thing is happening at x is equal to 3. At x equals 3, as we approach from the left, the slope looks like it is decreasing. It is approaching-- I don't know-- maybe around negative 1. But as we approach from the right it looks like the slope is 0. So we do not have the same limit of the secant slope as we approach from the left- and right-hand sides. So at both of these points we see the derivative jump, and it looks like f of x is not differentiable.